\gray
\section*{Introduction}
In this manuscript, we  try to follow the modern approach of algebraic geometry in viewing the varieties as an intrinsic geometrical object, by concentrating on the their intrinsic properties, rather than their extrinsic ones, that depends on their embedding in a specific affine or projective space.
\section{What Does Algebraic Geometry Study!?}
Geometry encoded by polynomials / regular functions.\\
Classification problem.\\
\tcb{Try to get to the closest answer}.
\section*{Algebraic Varieties}
\section{Zariski-Closed sets}
\subsection{Zariski-Closed sets in $\AF^n$}
\begin{definition}[The Affine Space $\AF^n$]
For any field $K$, $n\in\NN_{>0}$ we define the affine space $\AF^n=K^n$.
\end{definition}
\begin{definition}[Zariski-closed set on $\AF^n$]
We define the Zariski-closed set on $\AF^n$, to be the common zeros of a collection of polynomials in $K[x_1,x_2,...,x_n$.
\end{definition}
We find that the set of all Zariski-closed sets of $\AF^n$ define a topology, by closed sets, on $\AF^n$. Where, its closed sets are Zariski-closed sets.
\begin{definition}[Zariski Topology on $\AF^n$]
We call this topology the Zariski topology on $\AF^n$.
\end{definition}
\begin{definition}[Morphism of Zariski-closed sets in $\AF^n$]

\end{definition}
\tcb{justify the definition} and all other definitions.


\begin{fact}
\tcb{Although the Zariski topology of $\AF^2$ is not the product topology of $\AF^1\times \AF^1$. The affine space $\AF^2$ is isometric to $\AF^1\times \AF^1$ as a Zariski-closed of ??} \tcb{it does not make sense, does it?}.
\end{fact}
\begin{proof}
\end{proof}

\subsection{Morphisms of Zariski-Closed Sets in $\AF^n$}
\begin{theorem}[Pull-back of Zariski-closed Sets of affine spaces]
Let $V\subseteq \AF^n, W\subseteq \AF^m$ be Zariski-closed sets of \tcb{affine spaces}, and $F:V\rightarrow W$ a morphism of Zariski-closed sets. Then, the pull-back of $f$ is $F^{\#}:K[W]\rightarrow K[V]$ a morphism of $K$-algebras.
Moreover, $F^{\#}$ is injection iff $F$ is dominant, and $F^{\#}$ is surjection iff $F$ is an isomorphism from $V$ to $F(V)$.
\end{theorem}
\begin{proof}
\tcb{missing part}
\begin{itemize}
\item  Let $F$ be dominant, then $\VV(\II(F(V)))=\bar{F(V)}=W$, and let $g\in\ker{F^{\#}}$, then $gF=F^{\#}g=0_{K[V]}\Rightarrow g(F(V))=\{0_K\}\Rightarrow g\in \II(F(V))$. Hence, $g(W)=\{0_K\}\Rightarrow g\in\II(W)$, i.e. $g=0_{K[W]}$. I.e. $F^{\#}$ is injective.
\item Let $F^{\#}$ be injective. $\bar(F(V))\subseteq W$. Let $g\in \II(F(V)$, then $(F^{\#}g)(V)g(F(V))=\{0_K\}$. Hence, $g\in \ker{F^{\#}}\Rightarrow g=0_{K[W]}$, i.e. $g\in \II(W)$. Hence, $\II(F(V))\subseteq \II(W)$, then $W=\VV(\II(W))=\VV(\II(F(V)))=\bar{F(V)}$. I.e. $F^{\#}$ is dominant.
\item let $F$ be an isomorphism from $V$ to $F(V)$. \tcb{?}
\item let $F^{\#}$ be surjective. Then, $\forall g:V\rightarrow K, \exists g':W\rightarrow K$, such that $g=F^{\#}g'=gF$.
    Let $x,y\in V$ such that $F(x)=F(y)$, then $f(x)=gF(x)=gF(y)=f(y)$. Let $f_i(p)=P_l$, $i=1..n$ be the coordinate functions on $V$, then we find that $x_i=y_i, i=1..n$. I.e. $F$ is injection. Hence, it is a bijection to its image.\\
    \tcb{show that the inverse is a morphism itself}.

\end{itemize}


\end{proof}
\begin{example}
\end{example}


\begin{corollary}
Let $V\subseteq \AF^n, W\subseteq \AF^m$ be Zariski-closed sets of \tcb{affine spaces}, then $V$ and $W$ are isomorphic, as Zariski-closed sets of \tcb{affine spaces}, iff $K[V]$ and $K[W]$ are isomorphic, as $K$-algebras.
\end{corollary}
\begin{proof}

\end{proof}
\begin{example}
\end{example}

What if the isometry of the rings was not the pull-back??


Can we generalize the above to the case of projective spaces?




\begin{theorem}[Hilbert's Nullstellensatz Theorem]
\end{theorem}
\subsection{Zariski-Closed sets in $\PR^n$}
\subsection{Zariski closure}

\section{Quasi-Projective Variety}

The below definition of quasi-projective variety emphasis the intrinsic properties of varieties rather that their ambient space, which they are embedded in. Therefore, it encompasses both projective and affine varieties,\tcb{and others}, and pave the way to the definition of \tcb{algebraic variety}.

Can we define the quasi projective variety to be any \tcb{set} that can be embedded in a projective space $\PR^n$ as an intersection of Zariski-open and Zariski-closed sets. \tcb{(It does not really make sense, does it? How would we define the embedding in this case?)}

\begin{definition}[Quasi-Projective variety]
We define the quasi projective variety to be a locally closed set in a projective space.
\end{definition}


\begin{definition}[Morphism of Quasi-Projective variety]
??
\end{definition}

\begin{definition}[Affine Algebraic variety]
We call the quasi-projective variety an affine algebraic variety iff it is isomorphic as quasi-projective variety to a Zariski-closed set in some affine space $\AF^m$.
\end{definition}

\begin{definition}[Projective Algebraic variety]
We call the quasi-projective variety an projective algebraic variety iff it is isomorphic as quasi-projective variety to a Zariski-closed set in some projective space $\PR^m$.
\end{definition}
\begin{remark}
Every Zariski-closed set in some affine/ projective space is an affine/ projective algebraic variety. That, the identity map is an isomorphism of quasi-projective varieties.
\end{remark}


\section{Intersection Theory}
\begin{theorem}[B�zout's Theorem in $\PR^2$]
Having two curves in $\PR^2$, given by $\VV(f),\VV(g)$, for two polynomials $f,g$ of degrees $n,m$, respectively. Then, if $f$ and $g$ does not have common factors,  then the the intersection of $\VV(f),\VV(g)$ consists of nm points. And, these points are distinct, provided that the curves are not tangents for each other at any point.
\end{theorem}
\begin{proof}
\end{proof}
\section{Smoothness and Singularity}
\begin{theorem}[Hironaka's Desingularization Theorem]
Let $V$ be a quasi-projective variety. Then there exists a smooth quasi-projective variety. Then there exists a smooth quasi-projective variety $X$ and a projective birational morphism $X$.
\end{theorem}


\section{Maps to Projective Space}
\subsection{Vector and Line Bundles}
\begin{definition}\label{VectorBundle}
Let $X\subseteq\PR^n$ be a quasi-projective variety, we define the vector bundle of rank $m$ of $X$ to be the pair $(E,\pi)$, where $E$ is a quasi-projective variety, and $\pi:E\rightarrow X$ a morphism of quasi-projective varieties, such that there is a family $\{(U_i,\varphi_i),i\in I\}$, where $\{U_i,i\in I\}$ is a Zariski-open cover of $X$, and $\varphi_i:\pi^{-1}(U_i)\rightarrow U_i\times K^m$ are isomorphisms of quasi-projective varieties. such that:
\begin{enumerate}
  \item\label{VB1}The isomorphisms are fiber preserving, i.e. the following diagrams commute:
$\left.\xymatrix{
    \pi^{-1}(U_i) \ar[r]^{\varphi_i}\ar[dr]_{\pi}&U_i\times K^m\ar[d]^{P_l}\\
    & U_i}\right. ... \ast	$\\
    for $i\in I$, and $P_l$ the product projection on $U_i$.
  \item $\varphi_i$ are linearly compatible, that:
$\begin{array}{cccl}
\varphi_i\circ\varphi^{-1}_j:&U_i\bigcap U_j\times K^m &\longrightarrow &U_i\bigcap U_j\times K^m\\
&(p,v)&\longmapsto&\left(p,\tilde{\varphi}_i\varphi^{-1}_j(p,v) \right)
\end{array}$ is linear for each $p\in U_i\bigcap U_j$. Where $\tilde{\varphi}_i$ denotes the second argument of $\varphi_i$, i.e. its projection on $K^n$.
\end{enumerate}
\end{definition}
And we call $\pi^{-1}(x)$ the fiber of $x\in X$, and denoted by $E_x$.
\begin{fact}
The conditions of the previous definition are well-defined.
\end{fact}
\begin{proof}
Let $U\subseteq U_i$, then using \ref{VB1}, we find that\\
$\xymatrix{
\pi^{-1}(U_j) \ar[dr]_{\pi}&U\times K^m\ar[d]^{P_l}\ar[l]_{\varphi^{-1}_j}\\
 & U}$ commute. Hence, $\pi(\varphi^{-1}_j(U))\subseteq U\times K^m$.\\
$\xymatrix{
   \pi^{-1}(U) \ar[r]^{\varphi_i}\ar[dr]_{\pi}&U_i\times K^m\ar[d]^{P_l}\\
  & U}$ commute. Hence, $\varphi_i(\pi^{-1}(U))\subseteq U\times K^n$.\\
Then, $\varphi_i \varphi^{-1}_j$, defines a inner map on $U_i\bigcap U_j$.
\end{proof}
\begin{lemma}
The rank of vector bundles is well-defined.
\end{lemma}
\begin{proof}
Let $(E,\pi)$ be a vector bundle of $X$. And let $\{(U_i,\varphi_i):i\in I\}$ a local trivialization of $(E,\pi)$ of rank $n$, and $\{(V_i,\psi_i):i\in J\}$ a local trivialization of $(E,\pi)$ of rank $m$.\\
Then, since both of $\{U_i:i\in I\}$ and $\{V_i:i\in J\}$ covers $X$, Then, $U_{\alpha}\bigcap V_{\beta}\neq \empty$, for some $\alpha\in I, \beta\in J$. Then, $U_{\alpha}\bigcap V_{\beta}\times K^n\stackrel{\varphi_{\alpha}}{\cong}\pi^{-1}(U_{\alpha}\bigcap V_{\beta})\stackrel{\psi_{\beta}}{\cong}U_{\alpha}\bigcap V_{\beta}\times K^m$.
Hence, n=m. I.e. the rank of the vector bundle is an intrinsic property of the $(E,\pi)$ and does not depend on the considered local trivialization.
\end{proof}
\begin{proposition}
For a vector bundle $(E,\pi)$ of $X$, there is a maximal local trivialization.
\end{proposition}
\begin{proof}
Let $(E,\pi)$ be a vector bundle of $X$. And let $\{(U_i,\varphi_i):i\in I\}$ and $\{(V_i,\psi_i):i\in J\}$ be local trivializations of $(E,\pi)$. We will see that $\{(W_i,\theta_i):i\in I\sqcup J \}$ is a local trivialization of $(E,\pi)$, where $I\sqcup J$ is the disjoint union of $I$ and $J$, and $(W_i,\theta_i)=(U_i,\varphi_i)$ for $i\in I$ and $(W_i,\theta_i)=(V_i,\psi_i)$ for $i\in J$. That,\\
\begin{itemize}
  \item The previous lemma shows that both local trivializations have the same rank, let it be $n$.
  \item The first condition is satisfied for each element of either local trivializations, so it is satisfied for each element of the disjoint union.
  \item \tcb{proof the linear compatibility, if it is correct, or give a counterexample otherwise. If it is not correct, then we will end up with two different structures of $E_x$! Otherwise, if it is correct, that helps in proving the uniqueness of the $K$-vector structure on $X$, if it is so. Sho what it is}. Acrtually, it looks like that there is a maximal, but not maximum. Ex, the trivial vector bundle and identity morphism, and Id+constant! Double check it , and check its effect on the vector structure of $E_x$.
\end{itemize}
\end{proof}
\tcb{If the $K$-vector structure on $E_x$ is unique, that iluustrate it as an intrinsic property of the vector bundle like the rank}

\begin{proof}
Let $(E,\pi)$ be a vector bundle of $X$ of rank $m$ with respect to the local trivialization $\{(U_i,\varphi_i):i\in I\}$. Then, for any other local trivialization $(\{(V_i,\psi_i):i\in J\})$ that makes $(E,\pi)$ a vector bundle of $X$, it makes it so of rank $m$.\\
\tcb{proof}
\end{proof}
\begin{fact}
Let $(E,\pi)$ be a vector bundle of $X$ of rank $m$. Then, the fiber $E_x\pi^{-1}(x)$ is equipped with a well defined $K$-vector space of dimension $m$.
\end{fact}
\begin{proof}
Since $\{U_i:i\in I\}$ is a cover for $X$, then $\exists i\in I$, such that $x\in U_i$, then we define:\\
$v+_iw:=\varphi^{-1}_i\left(x,(\tilde{\varphi}_i(v)+\tilde{\varphi}_i(w))\right), \forall v,w\in E_x$\\
$\lambda \cdot_i v:=\varphi^{-1}_i\left(x,\lambda (\tilde{\varphi}_i(v))\right), \forall v\in E_x,\lambda \in K$\\
We notice that these operations are well-defined, i.e. does not depend on the choice of $U_i$ that contains $x$. That,:\\
\begin{itemize}
  \item $E_x\neq\empty$ that $\varphi^{-1}(x,0)\in E_x$.
  \item The operation $+_i$ is well defined on $E_x$, i.e. it does not depend on the choice of $U_i$. That, if $x\in U_i\bigcap U_j$ then, $\forall v,w\in E_x$\\
      $\varphi_i(v+_jw)=
      \varphi_i\varphi^{-1}_j(x,\tilde{\varphi}_j(v)+\tilde{\varphi}_j(w))\\=
      (x,\tilde{\varphi}_i\varphi^{-1}_j(x,\tilde{\varphi}_j(v)+\tilde{\varphi}_j(w)))\\=
      (x,\tilde{\varphi}_i\varphi^{-1}_j(x,\tilde{\varphi}_j(v))+\tilde{\varphi}_i\varphi^{-1}_j(x,\tilde{\varphi}_j(w)))\\=
      (x,\tilde{\varphi}_i\varphi^{-1}_j\varphi_j(v)+\tilde{\varphi}_i\varphi^{-1}_j\varphi_j(w))=
      (x,\tilde{\varphi}_i(v)+\tilde{\varphi}_i(w))$.\\Hence, $v+_iw=\varphi^{-1}_i(x,\tilde{\varphi}_i(v)+\tilde{\varphi}_i(w))=v+_jw$
  \item The operation $\cdot_i$ is well defined on $E_x$, i.e. it does not depend on the choice of $U_i$. That, if $x\in U_i\bigcap U_j$ then, $\forall v\in E_x,\lambda\in K$\\
      $\varphi_i(\lambda+_j v)=
      \varphi_i\varphi^{-1}_j(x,\lambda\cdot\tilde{\varphi}_j(v))=
      (x,\tilde{\varphi}_i\varphi^{-1}_j(x,\lambda\cdot\tilde{\varphi}_j(v)))\\=
      (x,\lambda\cdot\tilde{\varphi}_i\varphi^{-1}_j(x,\tilde{\varphi}_j(v)))=
      (x,\lambda\cdot\tilde{\varphi}_i\varphi^{-1}_j\varphi_j(v))=
      (x,\lambda\cdot\tilde{\varphi}_i(v))$. Hence,
      $\lambda \cdot_i v=\varphi^{-1}_i(x,\lambda\cdot\tilde{\varphi}_i(v))=\lambda \cdot_i w$
  \item Since $K^n$ is $K$-vector space, one can readily check that $(E_x,+_i,\cdot_i)$ is a $K$-vector space, and it is of dimension $n$. That, $\{\varphi^{-1}_i(e_{\alpha}):\alpha=1..n\}$ forms a basis for $E_x$ for $\{e_{\alpha}:\alpha=1..n\}$ a basis of $K^n$. \tcb{type it down when you have bit of time}.
\end{itemize}


\end{proof}
The above fact illustrate the importance of second condition in the above definition. Which allows us to equip $E_x$ with the $K$-vector space structure.\\
Whereas, the first condition assigns a copy of $K^n$ to each $x\in X$ and glue them together to form $E$.

\begin{fact}
The conditions of the Definition \ref{VectorBundle} are not redundant.
\end{fact}
\begin{proof}

\begin{enumerate}
  \item Give an example where only the first condition hold for what ever local trivialization one might choose.
  \item Give an example where only the second condition hold for what ever local trivialization one might choose.
  \item Otherwise, write the negation, and proof that one of the condition implies the other, or that the isomorphism property implies either one.
\end{enumerate}
\end{proof}

\tcb{Show how the vector bundle arise naturally.}

\begin{example}[The Trivial Bundle]
Let $X\subseteq \PR^n$ be a quasi-projective variety. We define the hyperplane bundle, of rank $m$, of $X$ to be $(E,\pi)$, such that:$E=X\times K^m$, $\pi:E\rightarrow X$ is the projection on $X$, i.e. $\pi((p,v))=p$
\end{example}
\begin{proof}
One can readily show that $\{(E,Id_E)\}$ satisfy, automatically, the definition of local trivialization of $X$. Actually it is even global trivialisation, which i turn explain the name.

Then, one readily find that $(E,\pi)$ is a vector bundle of rank $m$ of $X$.
\end{proof}

\begin{example}[The Tautological line Bundle]
Let $X\subseteq \PR^n$ be a quasi-projective variety. We define the hyperplane bundle of $X$ to be $(T,\pi)$, such that:\\
$T=\{(p,v), p\in X, v\in L_p \}$, where $L_p$ is the line in $K^{n+1}$ that goes through the origin and any representation of $p$, i.e. $L_p=\{t(p_0,p_1,...,p_n):t\in K\}$.\\
$\pi:T\rightarrow X$ is the projection on $X$, i.e. $\pi((p,v))=p$\\
Then $(T,\pi)$ becomes a line bundle.
\end{example}
\begin{proof}

Let $U_i=\{p\in X:P_l\neq 0\}$, $X=\displaystyle\bigcup_{i=0}^n U_i$. And, let\\
$\begin{array}{cccl}
\varphi_i:&\pi^{-1}(U_i)&\longrightarrow &U_i\times K\\
&(p,v)&\longmapsto&(p,v_i)
\end{array}$\\
\tcb{Show that this is the smallest local trivialization of $(H,\pi)$.}\\
For $i=0..n$,
\begin{itemize}
  \item $\varphi_i$ is well-defined. That,\\
  If $(p,v)=(q,w)\in \pi^{-1}(U_i)$, then $v_j=w_j$, for $j=0..n$, and $p=q$, i.e. $v,w\in L_p=L_q$. Then, $\varphi_i((p,v))=(p,v_i)=(q,w_i)=\varphi_i((q,w))$.
  \item $\varphi_i$ is a morphism. That, it is a composition of product map, and product projections, which in turn are morphisms.
  \item Let $\psi_i:U_i\times K\rightarrow \pi^{-1}(U_i)$ is defined by $\psi_i(p,k)=(p,kS(p)), \forall (p,k)\in U_i\times K$.\\
       Where,$S:\PR^n\rightarrow K^{n+1}$, given by\\ $S(p)=(P_0/{P_l},P_1/{P_l},...,P_{i-1}/{P_l},1,P_{i+1}/{P_l},...,P_n/{P_l})\in K^{n+1}, \forall p\in U_i$.\\
       $S$ \tcb{is well-defined}, and it is morphism of quasi-projective varieties that its components are rational functions of homogeneous polynomials of the same degree. Hence, $\psi_i$ is a morphism that it is the composition of product map, product projection and morphism.\\
       $\psi_i\varphi_i((p,v))=\psi_i((p,v_i))=(p,v_i S(p))$. Notice that, $v,S(p)\in L_p$, hence $v,v_iS(p)\in L_p$, i.e. $\exists \lambda\in K^{\ast}:v_j=\lambda (v_iS(p))_j$, for $j=0..n$. So, particularly, for $j=i$, we find that $v_i=\lambda (v_iS(p))_i=\lambda v_i\cdot 1=\lambda v_i$. Hence, $\lambda=1$ and $\psi_i\varphi_i((p,v))=(p,v_i S(p))=(p,v)\Rightarrow \psi_i\varphi_i=Id_{\pi^{-1}(U_i)}$.\\
       On the other hand, $\varphi_i\psi_i((p,k))=\varphi_i((p,k S(p)))=(p,(k S(p))_i)=(p,k)\Rightarrow \varphi_i\psi_i=Id_{U_i\times K}$.\\
       Therefore, $\varphi_i$ is an isomorphism, and its inverse is $\varphi^{-1}_i=\psi_i$.
       \item condition 1
       \item condition 2
\end{itemize}
\end{proof}

\begin{example}[The Canonical line Bundle]
\end{example}

\begin{example}[The Hyperplane Bundle]\label{HB}
Let $X\subseteq \PR^n$ be a quasi-projective variety. We define the hyperplane bundle of $X$ to be $(H,\pi)$, such that:\\
$H=\{(p,f|_{L_p}), p\in X, f \in(K^{n+1})^{\ast}\}$, where $L_p$ is the line in $K^{n+1}$ that goes through the origin and any representation of $p$.\\
$\pi:H\rightarrow X$ is the projection on $X$, i.e. $\pi((p,f))=p$\\
Then $(H,\pi)$ becomes a line bundle.
\end{example}
\begin{proof}
Let $U_i=\{p\in X:P_l\neq 0\}$, $X=\displaystyle\bigcup_{i=0}^n U_i$. And, let\\
$\begin{array}{cccl}
\varphi_i:&\pi^{-1}(U_i)&\longrightarrow &U_i\times K\\
&(p,f|_{L_p})&\longmapsto&\left(p,f|_{L_p}(S(p))\right)
\end{array}$\\
Where, $S(p)=(P_0/{P_l},P_1/{P_l},...,P_{i-1}/{P_l},1,P_{i+1}/{P_l},...,P_n/{P_l})\in K^{n+1}, \forall p\in U_i$.\\
\tcb{Show that this is the smallest local trivialization of $(H,\pi)$.}\\
For $i=0..n$,
\begin{itemize}
  \item $\varphi_i$ is well-defined. That,\\
  If $(p,f|_{L_p})=(q,g|_{L_q})\in \pi^{-1}(U_i)$, then $f|_{L_p}=g|_{L_q}$ and $p=q$ , i.e. $L_p=L_q$ and $S(p)=S(q)\in L_p$. Hence, $f|_{L_p}(S(p))=g|_{L_q}(S(q))$ and $\varphi_i\left((p,f|_{L_p})\right)=\varphi_i\left((q,g|_{L_q})\right)$.
  \item $\varphi_i$ is a morphism. That,\\
  \tcb{!?}
  \item $\varphi_i$ is surjective. That,\\$\forall (p,k)\in U_i\times K$, $(p,kx_i|_{L_p})\in \pi^{-1}$, and $\varphi_i\left((p,kx_i|_{L_p})\right)=(p,kx_i|_{L_p}(S(p)))=(p,k\cdot 1)=(p,k)$.
  \item $\varphi_i$ is injective. That,\\If $(p,f|_{L_p}),(q,g|_{L_q})\in \pi^{-1}(U_i)$ such that $\varphi_i((p,f|_{L_p}))=\varphi_i((q,g|_{L_q}))$.\\
      Then, $(p,f|_{L_p}(S(p)))=\left(q,g|_{L_q}(S(q))\right)$. Then, $p=q$, $L_p=L_q$, and $f|_{L_p}(S(p))=g|_{L_p}(S(p))$.
      Notice that $\forall x\in L_p, x=\lambda S(p)$, for some $\lambda\in K$. Then, $f|_{L_p}(x)=f|_{L_p}(\lambda S(p))=\lambda f|_{L_p}( S(p))=\lambda g|_{L_p}( S(p))= g|_{L_p}(\lambda S(p))=g|_{L_p}(x)$, i.e. $f|_{L_p}=g|_{L_p}$. Hence, $(p,f|_{L_p})=(q,g|_{L_q})$.
  \item $\varphi^{-1}_i:U_i\times K\rightarrow \pi^{-1}(U_i)$ is defined by $\varphi^{-1}_i(p,k)=(p,kx_i|_{L_p}), \forall (p,k)\in U_i\times K$. That,
  \item Alternatively, $\varphi_i$ accepts the inverse $\psi_i$, defined by ... .That,
  \item $\varphi^{-1}_i$ is a morphism. That,
  \item $\varphi_i\circ\varphi^{-1}_j$

\end{itemize}
\end{proof}
Investigating \cite[Exersice 8.5.1/3]{Smith00}, one readily realizes that the set of hyperplane subvarieties are divisors of the hyperplane bundle, which justify the name of the hyperplane bundle. Still, one needs to note that they are not necessary all the divisors are hyperplane, that some them might be of lower dimension.

Getting back to the tautological line bundle, we note that it does not help in defining any rational map from the variety to the projective space, and its only divisor is the variety itself. That in turn does not give any additional information about $X$, which justify the name of the tautological bundle.


The idea is that when two functions have the same restriction on a line in $K^{n+1}$, that does not mean the the section that they define are the same. So, event hough the vector bundle in this case is of dimension $1$, still, the vector space of global section is of dimension $n+1$. \tcb{prove it!}\\

Why do we care about the fact that $E_x$ is closed? Where do we use it? Prove that it is closed anyway. Well, we need it to be either closed or open, so it forms a variety itself, then the construction of $\varphi_i$ form an isomorphism to $K^n$.\tcb{rewrite it}.

Which understanding consists with the definition of dual bundle?
\subsection{Construction of Vector Bundles}
\tcb{Is there a natural way related to these settings rather than the pull-back?}
\begin{lemma}
Let $f:X\rightarrow Y$ a morphism of quasi-projective varieties and let $(E,\pi)$ be a vector bundle of rank $n$ on $Y$. Then, fibered product $X\times_Y E$ is a quasi algebraic variety.
\end{lemma}
\begin{proof}
The product of quasi-projective varieties is a quasi-projective variety, that the Zariski topology of $\PR^n\times \PR^m$ is contained in the Zariski topology of $\PR^{n+m}$, (they not equal though \cite[Exersice 1.2.2]{Smith00}).\\
One readily find that $f\times \pi:X\times E\rightarrow Y\times Y$ is continuous, that both $f$ and $\pi$ are. Let $\Delta_y=\{(y,y):y\in Y\}$, we see that $\Delta_y$ is closed in $Y\times Y$. That, it is given by $\Delta_Y=\VV(y_1-y_2)\subseteq Y\times Y$. Hence, $X\times_Y E=\{(x,e):x\in X,e\in E:f(x)=\pi(e)\}=(f\times \pi)^{-1}(\Delta_Y)$ is closed $X\times E$. Therefore, it is locally closed in the ambient projective space of $X\times E$, i.e. it is a quasi-projective variety.\\
\end{proof}

\begin{method}[Pull-back of Vector Bundle]\label{PB-VB}
\tcb{view it as a category pull-back}\\
Let $f:X\rightarrow Y$ a morphism of quasi-projective varieties and let $(E,\pi)$ be a vector bundle of rank $n$ on $Y$. Then, the pull-back of $(E,\pi)$ by $f$, defines a vector bundle of rank $n$ on $X$
\end{method}
\begin{proof}
Let $\pi':X\times_Y E\rightarrow X$ defined by $\pi'((x,e))=x$. So, it is a morphism (for being the restriction of product projection on the fibered product).\\
Let $\{(U_i,\varphi_i):i\in I\}$ be a local trivialization of $(E,\pi)$ of rank $n$. Then, $\{U'_i=f^{-1}(U_i):i\in I\}$ forms an open \tcb{(why?)} cover for $X$. And define the family of maps $\varphi'_i:\pi'^{-1}(U'_i)\rightarrow U'_i\times K^n$, for $i\in I$, defined by $\varphi'_i((x,e))=(x,\tilde{\varphi}_i(e)),\forall (x,e)\in \pi'^{-1}(U'_i)$.\\
We notice that, for $i\in I$:
\begin{itemize}
  \item $\varphi'_i$ is well-defined. That, $\varphi_i$ is well-defined and that\\
  $(x,e)\in \pi'^{-1}(U'_i)=\pi'^{-1}(f^{-1}(U_i))=\{(\xi,\epsilon)\in X\times_Y E, \xi\in f^{-1}(U_i)\}=\\\{(\xi,\epsilon): \pi(\epsilon)=f(\xi)\in U_i\}=\{(\xi,\epsilon)\in X\times_Y E, \epsilon\in \pi^{-1}(U_i)\}$. Hence, $e\in \pi^{-1}(U_i)=\dom(\varphi_i)$, and $\tilde{\varphi}_i(e)$ is well-defined.\\
  The verification of the well-definition of $\varphi'_i$, emphasize the naturality of the choice of defining the new space to be the fibered product.
  \item $\varphi'_i$ is an isomorphism of quasi algebraic varieties. That,
  \item Cond 1
  \item Cond 2
\end{itemize}

\tcb{justify calling it pull-back}, by showing that it is a categorical pull-back.
\end{proof}
The above method shows how we can construct a vector bundle for $X$, naturally, for each vector bundle of $Y$. Here, a natural question arise.
\begin{question}
In the setting of the above method, can we always express, up to isomorphism, as vector bundles of $X$, as pull-backs of vector bundles of $Y$?
\end{question}
\begin{proof}[Answer]
\end{proof}

\begin{fact}
The collection of vector bundle of the variety is an invariant of the variety. That is to say that it is an intrinsic property of the variety and independent of its embeddings.
\end{fact}
\begin{proof}
\end{proof}

Based on the above fact, a natural question arise.
\begin{question}
\tcb{Given a collection of vector bundles, can e always retrive the variety. It looks like a tautological question, that the variety is the codomain of each map, reframe it to be given only total spaces.}
\end{question}
\begin{proof}[Answer]
\end{proof}





\begin{method}[Product of Vector Bundles]
\end{method}

\begin{method}[Limit of Vector Bundles]
\end{method}

\begin{method}[Colimit of Vector Bundles!?]
\end{method}

\subsection{Sections of Vector Bundle}
\begin{definition}[Sections of Vector Bundle]
Let $(E,\pi)$ be a vector bundle of $X$, then we define the section on the open set $U\subseteq X$, \tcb{Why open?}, to be a morphism $s:U\rightarrow E$ such that $\pi\circ s=Id_U$, i.e. it embeds $U$ in $E$.\\
And we define the global sections to be the sections on $X$.
\end{definition}
The section $s$, on $U\subseteq X$, makes the following diagram commute:\\
$\xymatrix{
U\ar[rd]_{Id_U}\ar[r]^{s}&E\ar[d]|{\pi}\ar[r]^{\varphi_i}& U\bigcap U_i\times K^n\ar[ld]^{P_l}\\
&U&}$

For a local trivialization $\{(U_i,\varphi_i):i\in I\}$, we will denote $\tilde{s}_i=\tilde{\varphi}_i\circ s$.\\
Since the $K$-vector space structure of $(E_x,+_i,._i)$ is unique, we can always define the section $s:X\rightarrow E$, such that $s(x)=0_{(E_x,+_i,._i)}=\varphi^{-1}_i((x,0))$. That, it is a well-defined morphism of quasi-projective varieties \tcb{(prove it)} and $\pi(s(x))=x\forall x\in X$.\\
We will see later the importance of global sections in studying the properties of $X$, through embedding $X$ in $E$, and defining rational maps.\\
We denote the set of all sections on $U\subseteq X$, by $\bcE(U)$.
\begin{lemma}
$\bcE$ forms a sheaf of modules over the sheaf of rings $\bcO_X$.
\end{lemma}
\begin{proof}
For a vector bundle $(E,\pi)$ of $X$, then:\\
For each open subset $U\subseteq X$, the set $\bcE(U)$ is a $\bcO_X(U)$-module, that:
\begin{itemize}
\item $\bcE(U)\neq \empty$, because the zero section on $U$ is included in $\bcE(U)$.
  \item $\forall s,s'\in \bcE(U)$, then $s+s':U\rightarrow E$ such that $\forall x\in U, (s+s')(x):=s(x)+_{E_x}s'(x)\in E_x$ is a well-defined operation, because of the uniqueness of the vector structure on $E_x$. We notice that, $\forall x\in U, (s+s')(x)=\varphi^{-1}(x,\tilde{s}(x)+_{K^n}\tilde{s'}(x))$. Hence, it is a composition of morphism, which in turn is a morphism.\\
      Also, $\forall x\in U, \pi\circ(s+s')(x)=\pi(\varphi^{-1}(x,\tilde{s}(x)+_{K^n}\tilde{s'}(x)))=P_l((x,\tilde{s}(x)+_{K^n}\tilde{s'}(x)))=x\Rightarrow \pi\circ(s+s')=Id_U$. Therefore, $s+s'\in\bcE(U)$.
  \item $\forall s\in \bcE(U),f\in\bcO_X(U)$, then $f\cdot s:U\rightarrow E$ such that $\forall x\in U, (f\cdot s)(x):=f(x)._{E_x} s(x)\in E_x$ is a well-defined operation, because of the uniqueness of the vector structure on $E_x$. We notice that, $\forall x\in U, (f\cdot s)(x)=\varphi^{-1}(x,f(x)._{K^n}\tilde{s}(x))$. Hence, it is a composition of morphism, which in turn is a morphism.\\
      Also, $\forall x\in U, \pi\circ(f\cdot s)(x)=\pi(\varphi^{-1}(x,f(x)\cdot_{K^n}\tilde{s}(x)))=P_l((x,f(x)\cdot_{K^n}\tilde{s}(x)))=x\Rightarrow \pi\circ(f\cdot s)=Id_U$. Therefore, $f\cdot s\in\bcE(U)$.
  \item Since $(E_x,+_{E_x},._{E_x})$ is a $K$-vector field, and since $f(x)\in K,\forall x\in U,f\in \bcO_X(U)$, then one automatically finds that $(\bcE(U),+,0)$ is an $\bcO_X(U)$-module.
  \item Since $(E_x,+_{E_x},)$
\end{itemize}
Now, we note that:
\begin{enumerate}
  \item For any open subsets $U_1\subseteq U_2\subseteq X$, $s\in EE(U_2)$, then $s|_{U_1}:U_1\rightarrow E$ is a morphism. We know that $\pi\circ s=Id_{U_2}$, restricting to $U_1$, we find that $\pi\circ s|_{U_1}=(\pi\circ s)|_{U_1}=Id_{U_2}|_{U_1}=Id_{U_1}$. Hence, $s|_{U_1}\in \bcE(U_1)$. Then the map $|:\bcE(U_2)\rightarrow \bcE(U_1)$, defined by $|(s)=s|_{U_1}$, is a well-defined map, and it is a homomorphism of $\bcO_X(U_2)$-modules, that it is the restriction of the identity map, and that $\bcO_X(U_2)\subseteq \bcO_X(U_1)$.
  \item For the open subsets $U_1\subseteq U_2\subseteq U_3\subseteq X$, then the restriction of the identity map $\bcE(U_3)\rightarrow \bcE(U_1)$ is the composition of the restrictions of the identity maps $\bcE(U_2)\rightarrow \bcE(U_1)$ and $\bcE(U_3)\rightarrow \bcE(U_2)$.
  \item Let $\{U_{\alpha}:\alpha\in A\}$ be an open cover for $U\subseteq X$, and let $s_{\alpha}\in \bcE(U_{\alpha}),\forall \alpha\in A$ such that $s_{\alpha}|_{U_{\alpha}\bigcap U_{\beta}}=s_{\beta}|_{U_{\alpha}\bigcap U_{\beta}}$.\\
      Then we may define $s:U\rightarrow E$, such that, $\forall x\in U, s(x)=s_{\alpha}(x)$, where $x\in U_{\alpha}$.\\
      We notice that $s$ is well-defined, that if $x\in U_{\alpha}\bigcap U_{\beta}$, then $s(x)=s_{\alpha}(x)=s_{\beta}(x)$, i.e. it does not depend on the choice of the open set, of the above family, that contains $x$.\\
      $s|_{U_{\alpha}}=s_{\alpha}$ is a morphism, i.e. it pulls-back regular functions on $E$ to regular functions on $U_{\alpha}$. $\{U_{\alpha}:\alpha\in A\}$ be an open cover for $U\subseteq X$. Therefore, $\forall x\in U$, there is an open neighborhood, namely $U_{\alpha}$, such that $x\in U_{\alpha}$, which pulls-back regular functions on $E$ to regular functions on $U_{\alpha}$. I.e. $s$ is a morphism on $U$.\\
      $\forall x\in U,\exists \alpha\in A:x\in U_{\alpha}$, then $(\pi\circ s)(x)=\pi(s_{\alpha}(x))=x\Rightarrow \pi\circ s=Id_{U}$. Therefore, $s\in\bcE(U)$.
      Let $s'\in EE(U)$, such that $s'|_{U_{\alpha}}=s_{\alpha},\forall \alpha\in A$. Then $s'|_{U_{\alpha}}=s|_{U_{\alpha}}$, since $\{U_{\alpha}:\alpha\in A\}$ is a cover for $U$. Then $s=s'$, which prove the uniqueness of the such $s$.
\end{enumerate}
Which means that $\bcE$ forms a sheaf of modules over the sheaf of rings $\bcO_X$, called the sheaf of
\end{proof}
\tcb{Explain how one can make use of these sheaves}.
\begin{lemma}
$\bcE(X)$ forms a $K$-vector space.
\end{lemma}
\begin{proof}
\end{proof}
Recalling the above examples of vector bundle, we work out their sheaves of global sections.
\begin{lemma}[Local Sections of vector Bundle]\label{LS-VB}
Let $(E,\pi)$ a vector bundle of $X$, of rank $m$. And let $\{(U_i,\varphi_i):i\in I\}$ be a trivialization of $(E,\pi)$. Then, for any open set $U\subseteq X$, if $\exists i\in I: U\subseteq U_i$, then $\bcE(U)\cong\displaystyle\oplus_{i=1}^m\bcO_X(U)$.
\end{lemma}
\begin{proof}
Let $s\in\bcE(U)$, then $\forall x\in U, \varphi_i(s(x))=(x,(f_1(x),f_2(x),...,f_m(x)))\in U_i\times K^m$. $\varphi_i$ and $s$ are morphisms, therefor, the component function of their composition $f_1,f_2,...,f_n:U\rightarrow K$ are regular functions.\\
Then we can define the map $\mu_i:\bcE(U)\rightarrow \displaystyle\oplus_{i=1}^m\bcO_X(U)$ such that $\mu_i(s)=P_r(\varphi_i(s))=(f_1,f_2,...,f_m)$, as per the above settings, where $P_r$ is the product projection on the right argument.\\
We notice that $\mu_i$ is a homomorphism of $\bcO_X(U)$-modules, that:\\
$\forall f\in \bcO_X(U), s,s'\in \bcE(U)$, such that $\mu_i(s)=(f_1,f_2,...,f_m)$ and $\mu_i(s')=(f'_1,f'_2,...,f'_m)$, then:\\
$\forall x\in U, \mu_i(fs+s')(x)=P_r(\varphi_i(fs+s'))(x)
=P_r(\varphi_i(f(x)._{E_x}s(x)+_{E_x}s'(x)))\\
=P_r((x,f(x)(f_1(x),f_2(x),...,f_m(x))+(f'_1(x),f'_2(x),...,f'_m(x))))
=f(x)(f_1(x),f_2(x),...,f_m(x))+(f'_1(x),f'_2(x),...,f'_m(x))\\
=f(x)\mu_i(s)(x)+\mu_i(s')(x)$. I.e. $\mu_i(fs+s')=f\mu_i(s)+\mu_i(s')$.\\
Let $f_1,f_2,...,f_m\in \bcO_X(U)$, then we can define $s:U\rightarrow E$, such that $\forall x\in U,s(x)=\varphi^{-1}_i(x,(f_1(x),f_2(x),...,f_m(x)))$, since $\varphi^{-1}_i$ is a morphism, and the component functions are regular, then $s$ is a morphism. Also, we note that:\\ $\pi(s(x))=\pi(\varphi^{-1}_i(x,(f_1(x),f_2(x),...,f_m(x))))=P_l(x,(f_1(x),f_2(x),...,f_m(x)))=x=Id_U(x)\Rightarrow \pi\circ s=Id_U$. Hence, $s\in \bcE(U)$.\\
Then we can define the map $\nu_i:\displaystyle\oplus_{i=1}^m\bcO_X(U)\rightarrow \bcE(U)$ such that $\nu_i((f_1,f_2,...,f_n))=\varphi^{-1}_i((-,(f_1,f_2,...,f_n)))$.\\
We notice that $\nu_i$ is a homomorphism of $\bcO_X(U)$-modules, that:\\
$\forall f\in \bcO_X(U), \underline{f}=\{f_1,f_2,...,f_m\},\underline{f'}=\{f'_1,f'_2,...,f'_m\}\in \displaystyle\oplus_{i=1}^m\bcO_X(U)$, then:\\
$\nu_i(f\underline{f}+\underline{f'})(x)=\nu_i((f(f_1+f'_1),f(f_2+f'_2),...,f(f_n+f'_n)))(x)\\=
\varphi^{-1}_i(x,(f(x)(f_1(x)+f'_1(x)),f(x)(f_2(x)+f'_2(x)),...,f(x)(f_n(x)+f'_n(x))))\\=
f(x)._{E_x}\varphi^{-1}_i(x,(f_1(x),f_2(x),...,f_n(x)))+_{E_x}\varphi^{-1}_i(x,(f'_1(x),f'_2(x),...,f'_n(x)))\\=
(f\nu_i(\underline{f})+_{\bcE(U)}\nu_i(\underline{f}))(x)\Rightarrow
\nu_i(f\underline{f}+\underline{f'})=f\nu_i(\underline{f})+_{\bcE(U)}\nu_i(\underline{f})$.\\
Then, one can readily see that $\nu_i\circ\mu_i=Id_{\bcE(U)}$, and $\mu_i \circ \nu_i=Id_{\displaystyle\oplus_{i=1}^m\bcO_X(U)}$. Therefore, $\displaystyle\oplus_{i=1}^m\bcO_X(U)\cong\bcE(U)$.
\end{proof}
\begin{theorem}
There is a one-to-one correspondence between vector bundles of rank $n$ over a quasi-projective variety $X$ and the locally fee sheaves of modules of rank $n$ over $\bcO_X$.
\end{theorem}
\begin{proof}
We found in \ref{LS-VB} that the sheaf $\bcE$ is locally free sheaf of $\bcO_X$-modules, of the same rank of the considered vector bundle.\\
On the other hand, let $\bcE$ be a free sheaf of modules of rank $n$ over $\bcO_X$.\\
$\bcE$ is locally free, then there is an open cover of $X$, $\{(U_i):i\in I \}$, such that\\
$\bcE(U)\stackrel{\mu_i}{\cong}\displaystyle\oplus_{i=1}^n\bcO_X(U),\forall U\subseteq U_j$, for $j\in I$.\\
Let $E:=\{(x,(f_1,f_2,...,f_n)):\exists i\in I, x\in U_i:\exists s\in \bcE(U_i), \mu_i(s)=(f_1,f_2,...,f_n)\}$
Then we see that $(E,\pi)$ is a vector bundle for $\pi:E\rightarrow X$, defined by $\pi(x,(f_1,f_2,...,f_n))=x$. That, one may take the set $\{(U_i,\varphi_i):i\in I\}$ which is a local trivialization for $(E,\pi)$, for $\varphi_i:\pi^{-1}(U_i)\rightarrow U_i\times K^n$, defined by $\varphi_i((x,(f_1,f_2,...,f_n)))=(x,(f_1(x),f_2(x),...,f_n(x)))$, that, for $i\in I$:
\begin{itemize}
  \item $\varphi_i$ is well-defined. That,
  \item $\varphi_i$ is a isomorphism. That,
  \item 1 cond
  \item 2 cond
\end{itemize}
\tcb{prove it}.
\end{proof}
One should not confuse the the result in the previous example with sheaf of sections of open sets that are not contained in one of the component of some local trivialization, particulary, the global sections, as we will see in the following examples:
\begin{example}[Sections of Trivial Bundle]
Since $\{(X,Id_{X\times K^n}\}$ is a local trivialization for the trivial vector bundle of rank $n$. Then, based on \ref{LS-VB}, we find that $\bcE(U)\cong\displaystyle\oplus_{i=1}^n\bcO_X(U)$, for all open subsets $U\subseteq X$. Particularly, the sheaf of global sections is $\bcE(X)\cong\displaystyle\oplus_{i=1}^n\bcO_X(X)$.
\end{example}
\begin{example}[Sections of Tautological line Bundle]

\end{example}
\begin{example}[Sections of Canonical line Bundle]
\end{example}
\begin{example}[Sections of Hyperplane Bundle]
Let $(H,\pi)$ be the hyperplane bundle of $X\subseteq \PR^n$, example \ref{HB}. And let $g\in (K^{n+1})^{\ast}$.
Then we may define $s_g:X\rightarrow H$, such that $\forall p\in X:s_g(p)=(p,g|_{L_p})\in H$, which is readily shown to be a well-defined map. And, $\pi\circ s_g(p)=\pi((p,g|_{L_p}))=p=Id_X(p)\Rightarrow\pi\circ s_g=Id_X$.
Also, $s_g$ is a morphism of quasi-projective varieties, i.e. $s_g\in\bcE(X)$, that:\tcb{prove it!}\\
On the other hand, if $s:X\rightarrow H$, then $\forall p\in X, s(p)=(p,f_{s,p}|_{L_p})$, where each $f_{s,p}$ \tcb{now, based on the fact that $s$ is a morphism one should show that $f_{s,p}=f_{s,q},\forall p,q\in X$.}\\
Therefore, the global sections on $X$ are precisely those given, in the above setting, by functional in $(K^{n+1})^{\ast}$, i.e. by linear polynomials in $K[x_0,x_1,...,x_n]$.
\end{example}


\subsection{Line Bundle and Rational Map}
We study here the equivalence between rational maps to $\PR^n$ and pairs of line bundle and their global sections. This equivalence is in the sense that a rational map from a quasi-projective variety to the projective space $\PR^n$ is complectly determined by a line bundle and a global section. And that, every line bundle and its associated global section determine a rational map to $\PR$, as will see later on.
\begin{definition}[Linear Systems]
A nonzero vector space spanned by a set of global sections, associated with some line bundle $(L,\pi)$ of $X$, is called a linear system on $X$. And a linear system on $X$ is called complete linear system iff it is equal to $\bcE(X)$, and is denoted then by $|L|$.
\end{definition}
\tcb{show the importance/effect of having basis, i.e. independent set of elements}. One of which is to show that the sections do not all vanish at whole of $X$.
\begin{proposition}\label{LS-RM}
Let $\{_0s,_1s...,_ls\}$ be the basis of a linear system associated to $(L,\pi)$, a line bundle on $X$. then, the partial map $\xymatrix{\underline{s}:X\ar@{-->}[r]& \PR^l}$, defined by $x\mapsto [\tilde{_0s}_i(x):\tilde{_1s}_i(x):...:\tilde{_ls}_i(x)]$, taken for some trivialization, is a rational map.
\end{proposition}
\begin{proof}
$\underline{s}$ is well-defined. That it independent of the choice of the trivialization and the choice of the basis of the linear system.\\
\begin{itemize}
  \item \tcb{prove it}.
  \item
  \item
\end{itemize}
Since there is no guarantee that $\tilde{_js}_i$ do not all vanish at some point of $X$, then $\underline{s}$ is partially defined on $X$.\\
The restriction of $\underline{s}$ on the set $\{x\in X: \exists j=0..l \text{ such that } \tilde{_js}_i(x)\neq 0\}$ is a morphism of quasi-projective varieties. That,\tcb{prove it}.
\end{proof}
\begin{proposition}
Let $\xymatrix{F:X\ar@{-->}[r]& \PR^n}$ be a rational map, then $F$ is realize as $\underline{s}$, in the above setting for the pull-back of the hyperplane bundle $(H,\pi)$ of $\PR^n$, and the complete linear system spanned by the pull-back by the coordinate functions.
\end{proposition}
\begin{proof}
Let $U$ be the maximum open set in $X$, for which $F|_U$ is a morphism. Then, \ref{PB-VB} implies that that the pull-back of $(H,\pi)$, by $F|_U$ is a line bundle on $U$, given by $\pi':U\times_{\PR^n}H\rightarrow U$. We found that the functionals $_ix(p)=P_l$, $i=0..n$, form a basis for the complete linear system $\bcO_{\PR^n}(1)$. Then, their pull-back $_if(x)=F^{\#}(|ix)(x)=|ix((F(x)))$, $i=0..n$ form a basis for the complete linear system of the pull-back linear bundle, as shown in \tcb{ before Where?prove it as a lemma}.

\tcb{Then, we need to extend uniquely to $X$. For the section that could be done using inclusion}\\
\tcb{Then we need to show that the constructed is the same as $F$.}
\end{proof}

\begin{definition}[Base Locus]
We define the base locus of the linear system spanned by $\{_0s,_1s,...,_ns\}$ to be the subset of common zeros of $\{_0s,_1s,...,_ns\}$ in $X$.
\end{definition}
Note that by zero of $s$ in the above definition, we mean $x\in X:s(x)=0_{E_x}$.\\
\begin{fact}
The base locus of a linear system is independent of the choice of its base.
\end{fact}
\begin{proof}

\end{proof}
\begin{definition}[Free Base Locus]
We say that a linear system is base point free, if its base locus is empty.
\end{definition}
We may also say that the line bundle associated to the such linear system, is globally generated. That, the rational map introduced by \ref{LS-RM}, defines a morphism on $X$ \tcb{, so?}.
\begin{example}
\end{example}

\begin{definition}[Divisors]
The zero set of a nonzero global section $s$ of a line bundle $(L,\pi)$, is called a divisor.
\end{definition}
\section{Ideas}
\begin{problem}[Kakeya's Problem]
\end{problem}
\begin{theorem}[Pascal's Theorem (1639)]
\end{theorem}
\begin{theorem}[diver's Theorem]
\end{theorem}
\begin{conclusion}[Jacobian Conjecture]
\end{conclusion}
\section{Questions}
\begin{question}
Based on the argument at the beginning of \cite{Reid}, can we view each variety as a manifold? If yes, why do not we use its techniques in the study of algebraic varieties.
\end{question}
\black 